MAYBE

Initial complexity problem:
1:	T:
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(4, 0, 0, Ar_3, Ar_4, Ar_5))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f9(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5))
		(Comp: ?, Cost: 1)    f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_4, Fresh_5, Ar_4, Fresh_4))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_2, Fresh_3, Ar_4, Fresh_2))
		(Comp: ?, Cost: 1)    f6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_0, Fresh_1, Ar_4, Fresh_0))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 1:
	f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_4, Fresh_5, Ar_4, Fresh_4))
	f6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_0, Fresh_1, Ar_4, Fresh_0))
We thus obtain the following problem:
2:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_2, Fresh_3, Ar_4, Fresh_2))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f9(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5))
		(Comp: ?, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(4, 0, 0, Ar_3, Ar_4, Ar_5))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_2, Fresh_3, Ar_4, Fresh_2))
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f9(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5))
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(4, 0, 0, Ar_3, Ar_4, Ar_5))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

A polynomial rank function with
	Pol(f4) = 1
	Pol(f9) = 0
	Pol(f10) = 1
	Pol(koat_start) = 1
orients all transitions weakly and the transition
	f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f9(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5))
strictly and produces the following problem:
4:	T:
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_2, Fresh_3, Ar_4, Fresh_2))
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f9(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5))
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(4, 0, 0, Ar_3, Ar_4, Ar_5))
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
	start location:	koat_start
	leaf cost:	0

Applied AI with 'oct' on problem 4 to obtain the following invariants:
  For symbol f4: X_2 >= 0 /\ X_1 + X_2 - 4 >= 0 /\ -X_1 + X_2 + 4 >= 0 /\ -X_1 + 4 >= 0 /\ X_1 - 4 >= 0


This yielded the following problem:
5:	T:
		(Comp: 1, Cost: 0)    koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ]
		(Comp: 1, Cost: 1)    f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(4, 0, 0, Ar_3, Ar_4, Ar_5))
		(Comp: 1, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f9(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5)) [ Ar_1 >= 0 /\ Ar_0 + Ar_1 - 4 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 4 >= 0 /\ Ar_0 - 4 >= 0 ]
		(Comp: ?, Cost: 1)    f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1 + 1, Fresh_2, Fresh_3, Ar_4, Fresh_2)) [ Ar_1 >= 0 /\ Ar_0 + Ar_1 - 4 >= 0 /\ -Ar_0 + Ar_1 + 4 >= 0 /\ -Ar_0 + 4 >= 0 /\ Ar_0 - 4 >= 0 ]
	start location:	koat_start
	leaf cost:	0

Complexity upper bound ?

Time: 1.380 sec (SMT: 1.316 sec)